Chapter 7.3, Problem 1CP

Use the Collocation Method with $n=8$, $16$ to approximate solutions to the linear boundary value problems

a. $\{\begin{array}{c}{y^{\prime }}^{\prime }=y+\frac{2}{3}{e}^t\\ y\left(0\right)=0\\ y\left(1\right)=\frac{1}{3}e\end{array}$

b. $\{\begin{array}{c}{y^{\prime }}^{\prime }=\left(2+4t^2\right)y\\ y\left(0\right)=1\\ y\left(1\right)=e\end{array}$

Plot the approximate solutions together with the exact solutions (a)

$y\left(t\right)=t{e}^t/3$ and (b)

$y\left(t\right)=e^{t^2}$, and display the errors as a function of $t$ in a separate semilog plot.

To determine

To find: Approximate solution for linear boundary value problem by collocation method.

Explanation of Solution

% Program for Finite element solution of linear BVP
% Inputs are interval inter, boundary values bv, number of steps n
% Output: solution values c
function c=bvpfem(inter,bv,n)
a=inter(1);b=inter(2);ya=bv(1);yb=bv(2);
h=(b-a)/(n+1);
alpha=(8/3)*h+2/h; beta = (2/3)*h-1/h;
e=ones(n,1);
M=spdiags([beta*e alpha*e beta*e],-1:1,n,n);
d=zeros(n,1);
d(1)= -ya*beta;
d(n)= -yb*beta;
c=M\d;
c= bvpfem([0 1],[0 0.33e],8);

Collocation is applied $y^{″}=y+\frac{2}{3}{e}^t$ by expanding the solution into monomial basis functions $y(t)={\displaystyle \sum }_{j=1}^n{c}_j{t}^{j-1}.$ Substituting the boundary conditions gives $c_1=0and{c}_1+\dots +c_n=e/3$ he remaining n-2 equations use the differential equation:

  ${\displaystyle \sum }_{j=1}^n{c}_j\left[(j-1)(j-2)t_i^{j-3}-t_i^{j-1}\right]=\frac{2}{3}{e}^{t_i}$ for $i=2,\dots ,n-2$  Using evenly-spaced base points ti=(i-1)/(n-1) results in nn equations in the nn unknowns cj. After solving for cj by Gaussian elimination, the solution $y(t)={{\displaystyle \sum }}^{\text{}}{c}_j{t}^{j-1}$ is plotted on [0,1] for n=8 and 16, shown below. In addition, the differences between the approximate and exact solution are plotted for n=8. The differences for n=16 are near machine epsilon (not shown).

  

To determine

To find: Approximate solution for linear boundary value problem by collocation method.

Explanation of Solution

% Program for Finite element solution of linear BVP
% Inputs are interval inter, boundary values bv, number of steps n
% Output: solution values c
function c=bvpfem(inter,bv,n)
a=inter(1);b=inter(2);ya=bv(1);yb=bv(2);
h=(b-a)/(n+1);
alpha=(8/3)*h+2/h; beta = (2)*h-4/h;
e=ones(n,1);
M=spdiags([beta*e alpha*e beta*e],-1:1,n,n);
d=zeros(n,1);
d(1)= -ya*beta;
d(n)= -yb*beta;
c=M\d;

c= bvpfem([0 1],[1 e],16);

Similar to (a). The first and last equations are c1=1and c1+…+cn=e. The remaining n-2equations are ${\displaystyle \sum }_{j=1}^n{c}_j\left[(j-1)(j-2)t_i^{j-3}-2t_i^{j-1}-4t_i^2{t}_i^{j-1}\right]=0.$

Substituting  $t_i=\frac{\left(i-1\right)}{\left(n-1\right)} \text{for}i=2,\dots ,n-1$ and solving for  $c_j$  gives the approximate solutions shown below.